Oliveira

140 Reputation

2 Badges

7 years, 115 days

MaplePrimes Activity


These are replies submitted by Oliveira

Works well. Thanks again for the help acer.

I also thank you all for your attention.

I've also seen something similar to Dirichlet Conditions.

Regards,

Oliveira

This approach is more complicated. I will change the plate to a rectangular shape. Again, I am grateful for the information Rouben.

PDE := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

diff(diff(u(x, y), x), x)+diff(diff(u(x, y), y), y) = 0

(1)

BC := {u(0, y) = 20, u(2, y) = 100, u(x, 0) = 0, u(x, 1) = piecewise(x <= 1, 20, 100)}

 

pdsolve([PDE, op(BC)], u(x, y))

u(x, y) = Sum((160*sin((1/2)*n*Pi*x)*(cos((1/2)*Pi*n)-(5/4)*(-1)^n+1/4)*exp(-(1/2)*Pi*n*(y-9))-160*sin((1/2)*n*Pi*x)*(cos((1/2)*Pi*n)-(5/4)*(-1)^n+1/4)*exp(-(1/2)*Pi*n*(y-1))+160*sin((1/2)*n*Pi*x)*(cos((1/2)*Pi*n)-(5/4)*(-1)^n+1/4)*exp((1/2)*Pi*n*(y+1))-160*sin((1/2)*n*Pi*x)*(cos((1/2)*Pi*n)-(5/4)*(-1)^n+1/4)*exp((1/2)*Pi*n*(y+9))-40*sin(n*Pi*y)*((-1)^n-1)*(exp(Pi*n*(x+1))+5*exp(Pi*n*(x+2))-5*exp(Pi*n*(x+3))-exp(-Pi*n*(x-5))+exp(-Pi*n*(x-4))+5*exp(-Pi*n*(x-3))-5*exp(-Pi*n*(x-2))-exp(n*Pi*x)))/((-exp(5*Pi*n)+exp(4*Pi*n)+exp(Pi*n)-1)*n*Pi), n = 1 .. infinity)

(2)

``

In symbolic form, Maple returns a closed solution.

``

Download pde.mw

When I evaluate, Maple returns the following message:

Error, (in pdsolve/numeric) unable to handle elliptic PDEs.

Indeed, pdsolve cannot numerically solve elliptical pdes.

Thank you for your attention mmcdara. I will try to apply this procedure, because I also have little experience with Maple.

Thanks also to Rouben for the comment.

It's what I was needing. I am very grateful for your attention mmcdara.

Regards,

Oliveira

Sorry to bother you again acer. I didn't get the same result with the example below:

with(plots):

s1 := x^2 + y^2 + z^2 = 1;
s2 := x^2 - y^2 = z*(z - 1);
pon1 := intersectplot(surface(s1, x = -1 .. 1, y = -1 .. 1, z = -1 .. 1), surface(s2, x = -1 .. 1, y = -1 .. 1, z = -1 .. 1), axes = box, thickness = 4, orientation = [70, 40]);

temp1:=plottools:-transform((x,y,z)->[x,y,z])(pon1):
convert(plottools:-getdata(temp1)[3],listlist);

This what I was needing. I appreciate your attention acer.

Regards

Oliveira

Applying the procedure provided by Kotonum 19009, the coordinates do not appear in the form [x,y,z]. I would like the points to appear in the form of vectors or lists.

@Axel Vogt 

An attached document follows.
 

with(VectorCalculus)

``

This problem I took from a book, which asks to determine the magnetic field H inside a solid spherical cone.

 

             

 

The author solved the question using Ampere's circuit law and found the following expression:

 

                    

 

         

 

 

I tried to solve the issue using integration through FF expression:

  

                   

 

NULL

 

NULL

The coordinates of the infinitesimal volume element dV is:

 

rv = {p*cos(alpha), p*sin(alpha)*cos(phi), p*sin(alpha)*sin(phi)}

 

The coordinates of point P on the cone are:

 

rp = {0, r*cos(theta), r*sin(theta)}

 

Integration will be performed for:

 

"     0<p<1      0<alpha<Pi/(4)      0<phi<2 Pi    d(H)=((J)/(4 Pi r^(2))*(r)[u])dv  H=(&int;)[0]^(2Pi)(&int;)[0]^(Pi/(4))(&int;)[0]^(1)((J)/(4 Pi r^(2))*(r)[u])p^(2)sin(alpha) &DifferentialD;p &DifferentialD;alpha &DifferentialD;phi"
NULL

NULL

rR := `<,>`(p*sin(alpha)*cos(phi), p*sin(alpha)*sin(phi), p*cos(alpha))

Vector(3, {(1) = p*sin(alpha)*cos(phi), (2) = p*sin(alpha)*sin(phi), (3) = p*cos(alpha)})``

(1)

dA := `assuming`([simplify(sqrt(`&x`(diff(rR, phi), diff(rR, alpha)).`&x`(diff(rR, phi), diff(rR, alpha))))], [p > 0, alpha > 0, alpha < (1/4)*Pi])
NULL

p^2*sin(alpha)

(2)

AA := int(int(dA, alpha = 0 .. (1/4)*Pi), phi = 0 .. 2*Pi)

2*p^2*(1-(1/2)*2^(1/2))*Pi

(3)

rv := `<,>`(p*sin(alpha)*cos(phi), p*sin(alpha)*sin(phi), p*cos(alpha))

Vector(3, {(1) = p*sin(alpha)*cos(phi), (2) = p*sin(alpha)*sin(phi), (3) = p*cos(alpha)})

(4)

rp := `<,>`(r*sin(theta), 0, r*cos(theta))

Vector(3, {(1) = r*sin(theta), (2) = 0, (3) = r*cos(theta)})

(5)

rr := rp-rv

Vector(3, {(1) = r*sin(theta)-p*sin(alpha)*cos(phi), (2) = -p*sin(alpha)*sin(phi), (3) = r*cos(theta)-p*cos(alpha)})

(6)

ru := `assuming`([simplify(rr/sqrt(rr.rr))], [p > 0, alpha > 0, alpha < (1/4)*Pi])

Vector(3, {(1) = (r*sin(theta)-p*sin(alpha)*cos(phi))/sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2), (2) = -p*sin(alpha)*sin(phi)/sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2), (3) = (r*cos(theta)-p*cos(alpha))/sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)})

(7)

rJ := subs(J = i/AA, Typesetting[delayDotProduct](J, `<,>`(sin(alpha)*cos(phi), sin(alpha)*sin(phi), cos(alpha)), true))

Vector(3, {(1) = (1/2)*i*sin(alpha)*cos(phi)/(p^2*(1-(1/2)*sqrt(2))*Pi), (2) = (1/2)*i*sin(alpha)*sin(phi)/(p^2*(1-(1/2)*sqrt(2))*Pi), (3) = (1/2)*i*cos(alpha)/(p^2*(1-(1/2)*sqrt(2))*Pi)})

(8)

dH := `assuming`([simplify(subs(dv = p^2*sin(alpha), `&x`(rJ/(4*Pi(rr.rr)), ru)*dv))], [p > 0, alpha > 0, alpha < (1/4)*Pi])

Vector(3, {(1) = -(1/4)*sin(alpha)^2*i*sin(phi)*r*cos(theta)/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*(-2+sqrt(2))*Pi), (2) = -(1/4)*sin(alpha)*i*r*(-sin(alpha)*cos(phi)*cos(theta)+sin(theta)*cos(alpha))/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*(-2+sqrt(2))*Pi), (3) = (1/4)*sin(alpha)^2*i*sin(phi)*r*sin(theta)/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*(-2+sqrt(2))*Pi)})

(9)

dh1 := `assuming`([simplify(int(dH, p = 0 .. 1))], [alpha > 0, alpha < (1/4)*Pi, r > 0, r < 1, phi > 0, phi < 2*Pi, theta > 0, theta < (1/4)*Pi])

Vector(3, {(1) = -(1/4)*sin(alpha)^2*i*sin(phi)*r*cos(theta)*(int(1/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)), p = 0 .. 1))/((-2+sqrt(2))*Pi), (2) = -(1/4)*sin(alpha)*i*r*(-sin(alpha)*cos(phi)*cos(theta)+sin(theta)*cos(alpha))*(int(1/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)), p = 0 .. 1))/((-2+sqrt(2))*Pi), (3) = (1/4)*sin(alpha)^2*i*sin(phi)*r*sin(theta)*(int(1/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)), p = 0 .. 1))/((-2+sqrt(2))*Pi)})

(10)

``

Note that the result of the first integration is already quite complicated and I couldn't go any further.

``


 

Download Example1.mw

@acer 

Thanks for the fix acer.

Thank you for your attention mmcdara. In fact formal integration is complicated, so I'm going to try numerical integration.
Regards,
Oliveira

Thanks again for the tip.

Olveira

1 2 3 Page 1 of 3